Lesson 4 - adding forces, reducing a system of forces to a center point

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    created at April 11, 2021

    Lesson 4 - adding forces, reducing a system of forces to a center point

    Now you know that force is a vector quantity (as well as Torque). Several forces can be applied to mechanical system and it's usually a good step to simplify the system of forces, reducing them to one resulting force and one resulting torque. In order to do that we have to find a geometrical sum of force vectors.

    fig. 1.4.1 A system of forces and resulting force

    The system of four forces applied to the airplane:

    • Thrust force Ft
    • Gravity force Fg
    • air Drag force Fd
    • air Lift force Fl

    - replace them all with one equivalent force R

    There are two basic approaches how to find resulting vector. First one is graphical approach, also known as "parallelogram rule". For two vectors A and B, move one of them to make its origin coincident with second vector's origin. Then resulting vector R can be found as a diagonal of the parallelogram formed by these two vectors, that diagonal which comes out of the common vector origin (fig. 1.4.2).

    If we have more than two vectors, we can arrange them as a polygon line in such a way that each new vector comes out of the end of the previous one. Then we connect the start and end point of the polygon line - this is resulting vector R.

    fig. 1.4.2 Geometrical sum of vectors

    The benefit of graphical approach is that it gives you clear understanding of adding vectors. However, it's not always handy in practical use.

    The second approach is analytical one. A vector can be decomposed onto projections in a specific coordinate system. Thus vector can be presented as a set of numbers (projections). In three-dimensional space vector can be presented with three projections in following way:

    V⃗=Vx⋅i⃗+Vy⋅j⃗+Vz⋅k⃗\vec{V}=V_{x}\cdot \vec{i}+V_{y}\cdot \vec{j}+V_{z}\cdot \vec{k}
    fig. 1.4.3 Definition of vector using projections

    Vector V is expressed through unit vectors i, j, k, which are aligned along coordinate axis and have length of 1. Vx, Vy, Vz are vector's V projections onto axis x, y, z.

    Now it's very easy to find a resulting vector R. Each of its projections can be found as a sum of all according projections of initial vector system, i.e.:

    Rx=∑kFxk;Ry=∑kFyk;Rz=∑kFzk;R_{x}=\sum_{k}^{}F_{xk};\quad R_{y}=\sum_{k}^{}F_{yk};\quad R_{z}=\sum_{k}^{}F_{zk};

    For instance, there are two forces applied to a flying ball - air drag force:

    Fd⃗=−5⋅i⃗+2⋅j⃗\vec{Fd}=-5\cdot \vec{i}+2\cdot \vec{j}

    and force of gravity:

    Fg⃗=0⋅i⃗−3⋅j⃗\vec{Fg}=0\cdot \vec{i}-3\cdot \vec{j}
    fig. 1.4.4 Calculating resulting vector

    Projections of resulting vector R can be found as:

    Rx=Fdx+Fgx=−5+0=−5[N]R_{x}=Fd_{x}+Fg_{x}=-5+0=-5\:[N]
    Ry=Fdy+Fgy=2−3=−1[N]R_{y}=Fd_{y}+Fg_{y}=2-3=-1\:[N]

    And the vector R itself:

    R⃗=−5⋅i⃗−1⋅j⃗\vec{R}=-5\cdot \vec{i}-1\cdot \vec{j}

    (our problem is defined in XY plane).

    So, resulting vector is the one who can replace the whole system of vectors (forces or torques).

    Net Force

    Another important concept is a Net Force. Such force has equivalent impact on the body as initial system of loads. One can think that resulting force and net force are equal, but this is not always true. If impact lines of all applied forces are intersecting in one single point, then resulting force will be equal to net force. On the other hand, if applied forces do not intersect at one point - they will produce twisting effect on the body, thus replacing forces with resulting force will be not enough. In order to ensure equivalent loading scheme we have to apply twisting torque as well. So, in this case net force doesn't exists because we can't represent correct loading scheme with only one resulting force.

    fig. 1.4.5 Explaining resulting and net forces

    Probably the reader will become more familiar with this while solving problems. At this point all you need is keep in mind that net and resulting forces are not strictly the same.

    Sometimes it's convenient to parallel translate a force impact line to another position. For instance, it can be done to reduce all applied forces to one single point (center). The following rule exists: "Force F impact line can be translated to origin A, wherein the moment of force F around origin A should be applied in addition". After that translated force F' together with moment of force mA(F) provide equivalent loading:

    fig. 1.4.6 Reducing a force to a specific center

    Let's consider a man pulling the tire on the ice. He is applying the force not directly in the center but with some offset d. Thus twisting effect occurs and the tire both moves and twists.

    fig. 1.4.7 Example of reducing a force to a center

    Later the man began to pull the tire directly through the center. Now the tire just moves without twisting. Thus we have a different loading scheme with a different result. If we need to ensure the effect like in first case we have to add torque to the tire. So, someone else have to apply twisting torque to the tire while first on is pulling it. The value of this torque is force F times offset d.

    Conclusion

    Any system of applied loads can be reduced to resulting force and resulting torque vectors by using presented approaches.

    fig. 1.4.8 Resulting Force and resulting Torque vectors

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